当我运行程序时，“无”一直显示

Question

每当我运行我创建的计算器程序时，它都可以正常工作，但文本“ None”一直显示，我也不知道为什么。这是代码：

def add():

    print 'choose 2 numbers to add'

    a=input('add this')

    b=input('to this')

    print a+b

    return menu()

def sub():

    print 'choose 2 numbers to subract'

    a=input('subract this')

    b=input('from this')

    print b-a

    return menu()

def menu():

    print "hello, Welcome"

    print "these are your options"

    print "1. add"

    print "2. sub"

print menu()

loop=2

def sys():

    while loop==2:

        a=input("please choose")

        if a==1:

            print add()

        elif a==2:

            print sub()

        else:

            return menu(),sys()

print sys()

这是输出：

hello, Welcome

these are your options

1. add

2. sub

None    <-------------------------(this is what I'm talking about)

please choose

如果它对任何人都有帮助，那么这是我完成的计算器的代码（我粘贴它时看起来很混乱，但是当您复制和粘贴时它可以工作）

def add():

    print 'choose 2 numbers to add'

    a=input('add this')

    b=input('to this')

    print a+b

def sub():

    print 'choose 2 numbers to subract'

    a=input('subract this')

    b=input('from this')

    print b-a

def mul():

    print 'choose 2 numbers to multiply'

    a=input("multiply this")

    b=input("by this")

    print b*a

def div():

    print 'choose what numbers your want to divide'

    a=input('divide this')

    b=input('by this')

    print a/b

def exp():

    print 'choose your number you want to exponentiate'

    a=input('multiply this')

    b=input('by the power of this')

    print a**b

def menu():

    print "hello, Welcome"

    print "these are your options"

    print "1. add"

    print "2. sub"

    print "3. mul"

    print "4. div"

    print "5. expo"

    print "0. to end"

menu()

Answer 1

那是因为该函数menu()没有返回任何东西，默认情况下，python中的一个函数会返回None>>> def func():pass>>> print func()&nbsp; #use `print` only if you want to print the returned valueNone只需使用：menu() #no need of print as you're already printing inside the function body.从和sys()删除后的新版本。不必在每个函数内部使用，只需在其末尾调用该函数即可。return menu()add()sub()return menu()menu()while loopdef sys():&nbsp; &nbsp; while True:&nbsp; &nbsp; &nbsp; &nbsp; a = input("please choose")&nbsp; &nbsp; &nbsp; &nbsp; if a == 1:&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; add()&nbsp; &nbsp; # call add(), no need of print as you're printing inside add() itself&nbsp; &nbsp; &nbsp; &nbsp; elif a==2:&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; sub()&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; menu()&nbsp; &nbsp; &nbsp; &nbsp;# call menu() at the end of the loopwhile loop==2实际上loop==2首先计算表达式，如果是，True则while loop继续，否则立即中断。在您的情况下，因为您不更改loop变量的值，所以可以简单地使用while True。>>> loop = 2>>> loop == 2True